Calculating Kinetic Energy Using Uniform Velocity and Energy Equations

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Homework Help Overview

The discussion revolves around calculating kinetic energy in the context of a scenario where a box is lifted and then dropped. Participants are exploring the implications of work done by different forces, particularly focusing on the roles of gravity and the person lifting the box.

Discussion Character

  • Conceptual clarification, Assumption checking, Exploratory

Approaches and Questions Raised

  • Participants are questioning who is responsible for the work done in lifting the box and whether both the lifter and gravity contribute to the total work. There is also confusion about the calculation of kinetic energy after the box is dropped, with discussions on the relevant equations and reasoning behind them.

Discussion Status

Some participants have offered insights into the calculations and the nature of work done, while others express uncertainty and seek clarification. There is an ongoing exploration of different interpretations of the problem, particularly regarding the total work done and the assumptions involved in the scenario.

Contextual Notes

Participants are reminded of the homework rules that require them to make an effort to solve the problem independently. There is also a mention of the need for more research and understanding of the material relevant to the question.

UnknownQuestioner
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Homework Statement
Avdol lifted a 35 kg box at uniform velocity from the ground up to a height of 2.0m.

a) Jotaro commented Avdol did work. Polnareff says gravity did work. Who is correct? Explain?

b) If Avdol then accidentally dropped it. How much kinetic energy will there be right before it lands? Explain your reasoning. (2 marks)
Relevant Equations
Ek = 1/2 m v^2
For:
a) Avdol did the work because he is the force that is causing the displacement, right?

b) Is there another formula we would have to use? I am confused at how this would work out and what the answer would be.
 
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UnknownQuestioner said:
Homework Statement:: Avdol lifted a 35 kg box at uniform velocity from the ground up to a height of 2.0m.

a) Jotaro commented Avdol did work. Polnareff says gravity did work. Who is correct? Explain?

b) If Avdol then accidentally dropped it. How much kinetic energy will there be right before it lands? Explain your reasoning. (2 marks)
Relevant Equations:: Ek = 1/2 m v^2

For:
a) Avdol did the work because he is the force that is causing the displacement, right?

b) Is there another formula we would have to use? I am confused at how this would work out and what the answer would be.

That's correct for a). For b) where does the kinetic energy come from?
 
The kinetic energy would be the force of gravity pulling?
 
UnknownQuestioner said:
The kinetic energy would be the force of gravity pulling?

Can you think of something other than force?
 
Sorry I am not sure
 
UnknownQuestioner said:
Sorry I am not sure
Regarding this and your other post, the rules are that you must make an effort to solve the problem. We can help but we cannot do the work for you.

You need to have a think about the material you have been studying and what is relevant to a question like this.
 
Oh ok, sorry about that, I will re-attempt and try to do the problem again with more research. Thanks
 
UnknownQuestioner said:
a) Jotaro commented Avdol did work. Polnareff says gravity did work. Who is correct? Explain?

I think I can tell what the wording is implying, but this is an awful question? Both do work, and the total work is zero. Hmm
 
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I used √2gh = v to get velocity (6.3 m/s) then used that in Ek = 1/2 m v^2 to get a final kinetic energy of 690 J. Would this be the correct way to do this?
 
  • #10
etotheipi said:
I think I can tell what the wording is implying, but this is an awful question? Both do work, and the total work is zero. Hmm
Agreed, the intended answer is that both are right. But it is unclear whether the total work is zero. We are not told that Avdol stops it at that height, or whether we are to include the work done to accelerate it to the given speed.
 
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  • #11
UnknownQuestioner said:
I used √2gh = v to get velocity (6.3 m/s) then used that in Ek = 1/2 m v^2 to get a final kinetic energy of 690 J. Would this be the correct way to do this?
It is one way, but a bit roundabout. Combining those equations, Ek = 1/2 m v^2 = 1/2 m (√(2gh))^2 = mgh. How could you have got that straightaway?
 

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